Inaccessible set is the closure of its relative interior

Under [A1], use the dimension-general inputs in External input as follows. A curve with future timelike tangent almost everywhere is a causal absolutely continuous curve, so Minguzzi's arclength reparametrization applies. Its tangent is a positive multiple of the old tangent almost everywhere, hence remains future timelike. Conversely a locally Lipschitz timelike path is locally absolutely continuous. Thus the two conventions give the same chronological relation, and Chruściel's transitivity and openness apply to it. In dimension four the existing causal suppliers give the same conclusions with their choice hypotheses satisfied.

For $x\in B$ consider ${\gamma }_x(s)=(s,{\psi }_{s}x)$. Completeness keeps ${\psi }_{s}x$ in $C$ at every finite parameter. The flow equation for $-X$ and stationarity give ${\gamma }_x^{\mathrm{\prime }}=T-X=Nn$. Therefore $g({\gamma }_x^{\mathrm{\prime }},{\gamma }_x^{\mathrm{\prime }})=-N^2<0$ and ${\gamma }_x^{\mathrm{\prime }}$ is future directed. On each finite parameter interval the image is compact and the continuous lapse has a positive minimum, though no global uniform lower bound is required. In particular $(0,x)\ll (s,{\psi }_{s}x)$ for every $s>0$.

Write ${\mathrm{\Phi }}_a(t,x)=(t+a,x)$ for the complete flow of $T$. Since ${\mathcal{L}}_{T}g=0$, tensor-flow invariance gives ${\mathrm{\Phi }}_a^{\ast }g=g$ for every $a$. By [F6], these isometries preserve time orientation: for a future timelike vector $v$, $d{\mathrm{\Phi }}_a(v)$ remains timelike as $a$ varies continuously from zero, so cannot cross to the other component of the timelike cones. Applying ${\mathrm{\Phi }}_a$ and ${\mathrm{\Phi }}_{-a}$ to timelike curves, and using ${\mathrm{\Phi }}_a(U)=U$, gives ${\mathrm{\Phi }}_a(P)=P$ and ${\mathrm{\Phi }}_a(A)=A$ for $A=E\setminus P$. Openness of $P$ makes $A$ closed. The product identification ${\iota }_t:C\to \{t\}\times C$, ${\iota }_t(x)=(t,x)$, is a homeomorphism onto the slice with its subspace topology: its inverse is the restricted product projection. Hence $B={\iota }_0^{-1}(A)$ is closed in $C$. Finally, if $a\in A$ and $a\ll b\in P$, there is $u\in U$ with $b\ll u$; transitivity gives $a\ll u$, a contradiction. Thus $I_E^{+}(a)\subset A$ for $a\in A$.

By step 2.1, the open set $I_E^{+}((0,x))$ is contained in $A$. For $s>0$, its intersection with the slice $t=s$ contains $(s,{\psi }_{s}x)$ by step 1.2 and is relatively open. Pulling that intersection back by ${\iota }_s$, or equivalently translating by $-s$ and using ${\iota }_0$, gives an open neighborhood of ${\psi }_{s}x$ contained in $B$, by translation invariance of $A$. Thus ${\psi }_s(B)\subset {\mathrm{int}}_{C}B$. For $x\in B$, smoothness of the flow gives ${\psi }_{s}x\to x$ as $s\downarrow 0$. Every neighborhood of $x$ therefore meets ${\mathrm{int}}_{C}B$, so $x$ belongs to its closure: indeed no closed superset of that interior can omit $x$, since its open complement would then be such a neighborhood. This proves $B\subset {\overline{{\mathrm{int}}_{C}B}}^C$. Closedness of $B$ gives the reverse inclusion.